Mutual inductance of a long wire and a triangular loop?

I originally posted this to the advanced homework -section, but maybe this isn't advanced enough to warrant being posted there . If a mod sees this, feel free to delete the original on the other forum. I'd do it myself, but I can't.

1. The problem statement, all variables and given/known data

Calculate the mutual inductance for the long, straight conductor and a conducting loop shaped like an isosceles triangle in the picture below, when a = 10cm.

Magnetic flux: [itex]\phi = BA[/itex], where B is the magnetic flux density through a loop and A the area of the loop.

Magnetic flux density for a long straight wire(Ampere): [itex]B=\mu H = \frac{\mu I}{2 \pi R}[/itex], where R is the orthogonal distance from the wire.

3. The attempt at a solution

In order to calculate M, I need to find out the total flux through the triangular loop. We already know the expression for B, however dA is a bit of a mystery.

Looking at the picture, if at a distance r from the vertex of the triangle we mark:
\begin{cases}
\text{Height of top half of triangle} = y(r)\\
\text{Distance from the vertex} = r
\end{cases}
Then by similarity:
[tex]
\frac{1/2 a}{a} = \frac{y(r)}{r} \iff y(r) = \frac{r}{2}
[/tex]
and if we slice the triangle into small slices with height 2y and width dr:
[tex]
dA=2ydr = r dr
[/tex]
Now [tex]d\phi = d(B(r)A(r)) = B(r)dA(r)[/tex]

Therefore
\begin{array}{ll}
d\phi
&= \frac{\mu I r dr}{2 \pi (a+r)}\\
&= \frac{\mu I r dr}{2r \pi (\frac{a}{r}+1)}\\
&= \frac{\mu I dr}{2 \pi (\frac{a}{r}+1)}\\
&= \frac{\mu I dr}{2 \pi (\frac{a}{r}+1)}\\
&= \frac{\mu I}{2 \pi }\frac{dr}{(\frac{a}{r}+1)}
\end{array}
Looking at this it doesn't seem to be possible to integrate this expression with respect to r from 0 to a. My guess is my derivation of dA is wrong. Any help on that front?